Question

Question 3 (20 points) Given the following data with i = 15% L i on Machines EXY Investment (1) 30,000 AR AC N (years) $20,000 4000 10,000 4.400 000 14,000 8,600 10 Find the best alternative based on total investment, using: a) PW & repeatability on total investment. b) AW & repeatability on total investment. c) AW & coterminated on incremental investment.

economic

Answer 1

Answer a

For X

PW of Salvage Value = 4000/(1+15%)^5 = 4000/1.15^5 = 4000/2.0114 = $1988.71

Net Annual Benefit = 10000-4400 = $5600

PW of Annual Benefit = 5600*(1-(1+15%)^-5)/15%

= 5600*(1-1.15^-5)/0.15

= 5600*(1-0.4972)/0.15

=5600*0.5028/0.15

=$18772.07

Net PW = -20000+1988.71+18772.07= $760.78

Assuming repeatability, cashflow of 760.78 will reapt at end of 5 years PW of 760.78 = 760.78/(1+15%)^5 = 760.78/1.15^5 = 760.78/2.0114 = $378.23

Hence Total PW = 760.78+378.23=$1139.01

For Y

Net Annual Benefit = 14000-8600 = $5400

PW of Annual Benefit = 5400*(1-(1+15%)^-10)/15%

= 5400*(1-1.15^-10)/0.15

= 5400*(1-0.2472)/0.15

=5400*0.7528/0.15

=$27101.35

Net PW = -30000+27101.35= -$2898.65

Since X has a positive Net PW, it is best alternative

Answer b

For X

Net Initial Cost =First cost + PW of Salvage Value= -20000+1988.71 = -$-18011.29

AW of -18011.29, will be given by -18011.29 = A*(1-(1+15%)^-5)/15%

or -18011.29 = A*(1-1.15^-5)/0.15

or, -18011.29 = A*(1-0.4972)/0.15

or, -18011.29 = A*0.5028/0.15

or, A = -18011.29*0.15/0.5028

or, A =-5373.29

Hence Net AW = 10000-4400-5373.29=$226.71

For Y

AW of initial cost of $30000 is -30000=A*(1-(1+15%)^-10)/15%

or, -30000=A*(1-1.15^-10)/0.15

or, -30000=A*(1-0.2472)/0.15

or, -30000=A*0.7528/0.15

or, A = -30000*0.15/0.7528

=-$5977.68

Net AW = 14000-8600-5977.68=-$577.68

Since AW is positive for X it is best alternative

Answer c

AW of X is $226.71

Let's assume Y to be co terminated at end of year 5 and Salvage Value at end of year 5 be A.

AW of A will be given by, A = AW*((1+15%)^5-1)/15%

or, A = AW*((1.15)^5-1)/0.15

or, A = AW*(2.0114-1)/0.15

or, A = AW*((1.0114)/0.15

or, AW = A*0.15/1.0114 = 0.1483A

Hence Net AW of Y = -577.68+0.1483A

AW of X and Y will be equal when, 226.71= -577.68+0.1483A

or, 226.71+577.68=0.1483A

or, A = 804.39/0.1483 = 5424.07

It implies that if Salvage value of Y at end of year 5 is more than %5424.07 then Y is better Alternative else X is better alternative